Solve (2x-1)(2x+1)-2(x-2)(x+2)=

$$( 2 x - 1 ) ( 2 x + 1 ) - 2 ( x - 2 ) ( x + 2 ) =$$

Evaluate

$$2x^{2}+7$$

Solution Steps

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2x\right)^{2}-1^{2}-2\left(x-2\right)\left(x+2\right)$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-1^{2}-2\left(x-2\right)\left(x+2\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-1^{2}-2\left(x-2\right)\left(x+2\right)$$

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-1-2\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply -2 by x-2.

$$4x^{2}-1+\left(-2x+4\right)\left(x+2\right)$$

Apply the distributive property by multiplying each term of -2x+4 by each term of x+2.

$$4x^{2}-1-2x^{2}-4x+4x+8$$

Combine -4x and 4x to get 0.

$$4x^{2}-1-2x^{2}+8$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-1+8$$

Add -1 and 8 to get 7.

$$2x^{2}+7$$

Expand

$$2x^{2}+7$$

Solution Steps

Consider \left(2x-1\right)\left(2x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(2x\right)^{2}-1^{2}-2\left(x-2\right)\left(x+2\right)$$

Expand \left(2x\right)^{2}.

$$2^{2}x^{2}-1^{2}-2\left(x-2\right)\left(x+2\right)$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-1^{2}-2\left(x-2\right)\left(x+2\right)$$

Calculate 1 to the power of 2 and get 1.

$$4x^{2}-1-2\left(x-2\right)\left(x+2\right)$$

Use the distributive property to multiply -2 by x-2.

$$4x^{2}-1+\left(-2x+4\right)\left(x+2\right)$$

Apply the distributive property by multiplying each term of -2x+4 by each term of x+2.

$$4x^{2}-1-2x^{2}-4x+4x+8$$

Combine -4x and 4x to get 0.

$$4x^{2}-1-2x^{2}+8$$

Combine 4x^{2} and -2x^{2} to get 2x^{2}.

$$2x^{2}-1+8$$

Add -1 and 8 to get 7.

$$2x^{2}+7$$